A sound wave travels through a medium with a speed of 360 meters per second plus or minus nine percent. The wavelength of the sound wave is two meters plus or minus two percent. What is the percent uncertainty in the frequency of the sound wave?
Here we have a sound wave traveling along with a speed we’ll call 𝑣 and having a wavelength that we’ll call 𝜆. Both the wave speed and the wavelength have some uncertainty associated with them. Notice that these uncertainties are expressed as percents. Our question asks us to solve for the percent uncertainty in the frequency of the sound wave.
Let’s recall that when it comes to waves in general, and this includes sound waves, the speed of a wave is equal to its frequency times its wavelength. Dividing both sides of the equation by the wavelength 𝜆 so that it cancels out on the right, we see that wave frequency 𝑓 equals wave speed 𝑣 divided by wavelength 𝜆. In our scenario then, the frequency of our wave is 360 meters per second plus or minus nine percent divided by two meters plus or minus two percent.
We see that this is an instance of a number with some uncertainty being divided by another number also with some uncertainty. Whenever this happens, we find the resultant value, in this case the frequency of our wave, by carrying out the division we might expect, 360 meters per second divided by two meters, that’s equal to 180 inverse seconds or equivalently 180 hertz, plus or minus the sum of these two percent uncertainties. That is, we take the percent uncertainty in our numerator, nine percent, and we add it to the percent uncertainty in our denominator, two percent. Nine percent plus two percent is 11 percent. This is the percent uncertainty in the frequency of the sound wave.